9
$\begingroup$

Let $K/F$ be Galois with $G=Gal(K/F)$ and let $L$ be an intermediate field. Let $N\subseteq K$ be the normal closure of $L/F$. If $H=Gal(K/L)$ show that $Gal(K/N)=\cap_{\sigma\in G}\sigma H\sigma^{-1}$. (Exercise $8$, page $60$, Field and Galois Theory, Patrick Morandi.)

Help me a hint to prove it. Thanks a lot.

$\endgroup$
  • 2
    $\begingroup$ Show that $Aut_{\sigma(L)}(K) = \sigma Aut_L(K)\sigma^{-1}$ $\endgroup$ – Prahlad Vaidyanathan Nov 18 '13 at 9:30
  • $\begingroup$ Can you give me more details? $\endgroup$ – user109584 Nov 18 '13 at 9:34
  • $\begingroup$ Sorry I forgot to tag. $\endgroup$ – user109584 Nov 18 '13 at 9:48
  • $\begingroup$ Help me. please $\endgroup$ – user109584 Nov 19 '13 at 7:31
  • $\begingroup$ Have you been able to prove the hint I provided? $\endgroup$ – Prahlad Vaidyanathan Nov 19 '13 at 9:07
3
$\begingroup$

Here is my proof.

For convenience, let $M$ denote the group $\cap_{\sigma\in G} \sigma H\sigma^{-1}$. The notation $M(a)=a$ means that $a$ is fixed under the action of each element in $M$.

First we prove that $Gal(K/ N) \supseteq M$. It suffices to prove that $N$ is fixed by the group $M$. For any $a\in N$, since $N$ is the normal closure of $L/F$, there exists some $\sigma\in G$ such that $\sigma^{-1} (a) \in L$. Hence $H\sigma^{-1}(a)=\sigma^{-1}(a)$ because $L$ is fixed by $H$. Therefore, we see that $\sigma H\sigma^{-1}(a) = a,$ hence $M(a)=a$ because $M$ is contained in $\sigma H\sigma^{-1}$ by definition. Note that $a$ is arbitrary in $N$, hence $N$ is fixed by $M$.

Next we prove that $Gal(K/ N) \subseteq M$. Suppose $\tau\in Gal(K/N)$, for any $\sigma\in G$, $\tau\in\sigma H\sigma^{-1}\Leftrightarrow \sigma^{-1}\tau\sigma\in H\Leftrightarrow \sigma^{-1}\tau\sigma$ fixes $L$. So we only need to prove that $\sigma^{-1}\tau\sigma$ fixes $L$. $\forall a\in L$, we have $\sigma(a)\in N$ since $N$ is the normal closure of $L/F$. Because $\tau$ fixes $N$, we deduce that $\sigma^{-1}\tau\sigma(a) = \sigma^{-1}\sigma(a)=a$, which means that $\sigma^{-1}\tau\sigma$ fixes $L$, or $L$ is fixed by $\sigma^{-1}\tau\sigma$. We are done.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.